ĐKXĐ: \(-\sqrt{10}\le x\le\sqrt{10}\)
\(\left(x+4\right)\sqrt{10-x^2}=\left(x+4\right)\left(x-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\Rightarrow x=-4< -\sqrt{10}\left(l\right)\\\sqrt{10-x^2}=x-2\left(1\right)\end{matrix}\right.\)
Xét (1) với \(x\ge2\)
\(\Leftrightarrow10-x^2=\left(x-2\right)^2\)
\(\Leftrightarrow10-x^2=x^2-4x+4\)
\(\Leftrightarrow2x^2-4x-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1< 2\left(l\right)\\x=3\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=3\)